Problem: Let $h(x)=\dfrac{1}{x^2}+\dfrac{1}{x}+x$. $h'(x)=$
Answer: The strategy We can first rewrite each rational term of $h$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} h(x)&=\dfrac{1}{x^2}+\dfrac{1}{x}+x \\\\ &=x^{-2}+x^{-1}+x \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}(x^{-2}+x^{-1}+x) \\\\ &=\dfrac{d}{dx}(x^{-2})+\dfrac{d}{dx}(x^{-1})+\dfrac{d}{dx}(x) \\\\ &=-2x^{-3}+(-1)x^{-2}+1x^0 \\\\ &=-\dfrac{2}{x^3}-\dfrac{1}{x^2}+1 \end{aligned}$ In conclusion, $h'(x)=-\dfrac{2}{x^3}-\dfrac{1}{x^2}+1$.